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A twist is an automorphism $\theta$ of the identity functor of a monoidal category with braiding $c$, such that $\theta_{X \otimes Y} = c_{Y,X} c_{X,Y} (\theta_X \otimes \theta_Y)$. A braided monoidal category with twist is called balanced monoidal (not to be confused with a different notion of balanced).

It is known that modular categories are ribbon fusion, and thus have a twist. So all Drinfeld centers of fusion categories are balanced monoidal. Furthermore, Drinfeld centers of pivotal categories are pivotal again (see Figure 2 in this article. Pivotal structures and braidings give a balanced structure again. (Although I wouldn't know or care for now if they're ribbon.)

Are Drinfeld centers always balanced monoidal? If not, under which conditions are they balanced monoidal? If yes, how is the twist constructed?

Note that Drinfeld centers of monoidal categories without duals for every object may not have all duals either, and therefore needn't be pivotal. They can nevertheless be balanced monoidal. For example, take the category of all (complex) vector spaces. Infinite dimensional vector spaces don't have dual objects, so this category does not allow a pivotal structure. I'm strongly suspecting that its Drinfeld center is vector spaces again (although I don't know a rigorous proof, since it's not fusion), which has a trivial twist.

Edit: Drinfeld centers of rigid categories are always rigid (i.e. have all duals). Now since in a rigid braided category, twists are equivalent to pivotal structures. So a non-pivotal rigid category might potentially have a non-balanced Drinfeld center (although this is not certain, it might still "accidentally" have a balance if all nonpivotal objects for some reason don't have a half-braiding). I'm quite sure there are non-pivotal rigid categories (by an open conjecture, they'd need to be non-fusion, though) -- although I can't think of any right now -- so here is a place to look for counterexamples.

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  • $\begingroup$ In the toy example of an abelian group this, I believe, corresponds to finding, for a bilinear form $c$, a function $\theta$ such that $\theta(x+y)=c(x,y)+c(y,x)+\theta(x)+\theta(y)$. I believe in characteristic 2 there are examples of $c$ for which this is impossible. $\endgroup$
    – მამუკა ჯიბლაძე
    Sep 2, 2016 at 9:50
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    $\begingroup$ @მამუკა ჯიბლაძე, that sounds interesting. But I have the feeling you're looking for braided categories without a balance, not Drinfeld centers. How does an abelian group give rise to a Drinfeld center? When you take the Drinfeld center of $G$-graded vector spaces for some group $G$, you don't get $Z(G)$-graded vector spaces. $\endgroup$
    – Manuel Bärenz
    Sep 2, 2016 at 13:13
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    $\begingroup$ The chart on page 11 of my paper arxiv.org/pdf/1509.02937v1.pdf contains information that you might find useful. $\endgroup$
    – André Henriques
    Sep 3, 2016 at 23:04
  • $\begingroup$ Acknowledging that I missed the essential point, again in that toy example passing to the center means considering the category of pairs $\langle a,(\varphi^a_x)_x\rangle$ (the $\varphi^a_x$ interpreted as isomorphisms $a+x\to x+a$), with the monoidal structure $\langle a,(\varphi^a_x)_x\rangle+\langle b,(\varphi^b_x)_x\rangle=\langle a+b,(\varphi^a_{b+x}+\varphi^b_{x+a})_x\rangle$ and braiding $c(\langle a,\varphi^a\rangle,\langle b,\varphi^b\rangle)=(\varphi^a_b+x,x+\varphi^a_b)_x$; one might then figure out what precisely existence of $\theta$ means in this case. $\endgroup$
    – მამუკა ჯიბლაძე
    Sep 4, 2016 at 10:26
  • $\begingroup$ @AndréHenriques, thanks, I edited my question with a reference to your article. In terms of your Figure 2, my question is: "When can the arrow $\mathcal{Z}$ from 'tensor' to 'braided' be extended to 'balanced'?" or maybe "Is the image of the arrow $\mathcal{Z}$ inside the image of the forgetful arrow from 'balanced' to 'braided'?" My hope is "always"/"yes", but who knows. $\endgroup$
    – Manuel Bärenz
    Sep 4, 2016 at 12:05

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